0.35 grams of an unknown solid containing magnesium hydroxide is combined with 50 mL of 0.25M nitric acid. If 34.5mL of 0.1 NaOH was required to reach an endpoint, determine the concentration of magnesium hydroxide inthe sample.
Frankly, I don't think the problem is worded very well. This is an example of a problem in which the unknown amount of Mg(OH)2 is treated with a known amount of HNO3, which happens to be an excess. Then we determine the excess, subtract from the initial amount HNO3 there, and the difference is the amount of Mg(OH)2 that reacted.(Of course, the idea may have been that you were supposed to recognize the kind of problem you have.)
2HNO3 + Mg(OH)2 ==> Mg(NO3)2 + 2H2O
How much HNO3 was added initially. It is M x L = 0.25M x 0.050L = 0.0125 moles.
How much was in excess? It required 34.5 mL of 0.1 M NaOH; therefore, the NaOH neutralized M x L = 0.1M x 0.0345 L = 0.00345 moles HNO3.
Excess HNO3 = 0.0125 moles - 0.00345 moles = 0.00905 moles.
How many moles must have reacted with the HNO3. Just half as many (look at the coefficients in the balanced equation) so 0.00905/2 = 0.004525 moles.
g Mg(OH)2 = mols x molar mass = ??grams.
% Mg(OH)2 in the sample = (mass Mg(OH)2/mass sample)*100 = ??
Post your work if you get stuck.
I got it. Thanks a lot!
To determine the concentration of magnesium hydroxide in the sample, we need to follow these steps:
Step 1: Write the balanced equation for the reaction:
Mg(OH)2 + 2HNO3 -> Mg(NO3)2 + 2H2O
Step 2: Calculate the moles of nitric acid (HNO3):
Molarity (M) = moles of solute / volume of solution (L)
Moles of HNO3 = Molarity x Volume = 0.25 mol/L x 0.050 L = 0.0125 moles
Step 3: Determine the limiting reactant:
The balanced equation shows that the mole ratio between magnesium hydroxide (Mg(OH)2) and nitric acid (HNO3) is 1:2. Therefore, the moles of magnesium hydroxide can be calculated as:
Moles of Mg(OH)2 = 0.0125 moles x (1 mole Mg(OH)2 / 2 moles HNO3) = 0.00625 moles Mg(OH)2
Step 4: Calculate the concentration of magnesium hydroxide:
Concentration (M) = moles of solute / volume of solution (L)
Concentration of Mg(OH)2 = Moles of Mg(OH)2 / Volume of solution (L)
Volume of solution = Volume of NaOH used - Volume of HNO3 used
= 34.5 mL - (50 mL - 34.5 mL) = 19.0 mL = 0.019 L
Concentration of Mg(OH)2 = 0.00625 moles / 0.019 L = 0.329 M
Therefore, the concentration of magnesium hydroxide in the sample is 0.329 M.